Equilibria, Energetics and Elements (F325) - June 2011 Exam.

Scroll to see replies

Does anyone know of any good revision sites for OCR? Mainly with practice questions?? I know pastpaper. com..any others?

Reply 681

V.sey

Hey. This is a acid,base buffer Q

Which is in excess? NaOH or H2SO4? N.B- the actual moles are equal (working done with given conc and volume)

2NaOH + H2SO4 ----------> Na2SO4 + H2O

I do notice the 2: 1 ratio and the correct answer is H2SO4 in excess...can someone explain why?

Reply 682

Vinchenko

Original post by V.sey

Because if your base used up all the acid, then there wouldn't be any dissociated H+ protons kicking around to react with OH- ions when added, to minimise the change in pH - which is obviously the entire point of having a buffer :smile:

Reply 683

Vinchenko

Original post by apo1324

There are 4 precipitation reactions involving transtion metals which we have to know the colour changes for (and they are shown in the book), I think. :smile:

Cobalt, Fe2+, Fe3+ & Cu2+ with NaOH (aq)

Pink soln ---> blue ppt
Pale green soln ---> green ppt
Pale yellow soln ---> rusty brown ppt
Blue soln --->blue ppt

Just in case anyone else had the same question! Thanks, although damn you for making me get out a textbook at...10 to midnight! :wink:

Reply 684

ChubbyRain

Original post by sportycricketer

[CuCl4]2- is tetrahedral :smile:

Original post by student777

Tetrahedral ones are 4 ligands spread all around the TM ion. Bond angles are 109 degrees.

Square planar is like octahedral, but with the top and bottom ligands missing. There are 4 ligands, bond angles are 90 degrees. :smile:

thank you

Reply 685

INeedToRevise

Original post by wilsea05

if you write out the dissociation of Ca(OH)2 it will be
Ca(OH)2 -> Ca2+ + 2OH-
so its 1 mole of Ca(OH)2 : 2 moles of OH-
so you times the concentration by 2

Does that mean Fe(OH)3 will be 3moles of OH? So you multiply by 3?
This could be a sneaky question.

Reply 686

goldlock

Original post by RyuHADOUKEN

Just posted the answer guys ^ :smile:

Hahaha no problemo! :redface:

Thanks a lot mate :smile:

you think you could look at the question on that paper about the concentration of phenol's conjugate base in the buffer solution? I get 0.013 mol per dm cubed but the mark scheme says 0.13 :s-smilie:

Reply 687

BeekieBoo

Original post by hey-hey-hey

Does anyone know of any good revision sites for OCR? Mainly with practice questions?? I know pastpaper. com..any others?

http://www.knockhardy.org.uk/sci.htm This was given out earlier in the thread, it doesn't have any questions on as far as I know, but it's really good for notes, if that's any help.

Reply 688

BeekieBoo

Original post by goldlock

Thanks a lot mate :smile:

you think you could look at the question on that paper about the concentration of phenol's conjugate base in the buffer solution? I get 0.013 mol per dm cubed but the mark scheme says 0.13 :s-smilie:

That's exactly what I got, I also divided the rate question by 2 ... but I guess it makes sense to divide it by 4. I think it may just be a mistake on the mark scheme because even with the method they gave it still gives me 0.013 ...

Reply 689

mauvetard

Original post by student777

For Atoms, Bonds and Groups, first time round I got a C, retook and got a high A. I basically briefly learnt all the content then did past papers until I literally got sick at the sight of them! You do a few when you start so you know your weak areas. Then go over contennt you keep getting wrong. Then practice exam technique. Practise papers are manna from heaven for me! :P

Though it helped that it was a retake so I already knew most of it, and it's so easy compared to other A2 stuff.

Hope this helps :smile:

Don't ask what i'm doing up at this hour... xD

I might adopt your revision strategy for a change, it sounds like it'll work a hell lot for me. Thanks for replying and try to go to bed early next time missy :smile:

Reply 690

Wor Carroll

Original post by goldlock

Thanks a lot mate :smile:

you think you could look at the question on that paper about the concentration of phenol's conjugate base in the buffer solution? I get 0.013 mol per dm cubed but the mark scheme says 0.13 :s-smilie:

Original post by BeekieBoo

It is 0.013 because my teacher got that as well, so the mark scheme must be wrong :smile:

Reply 691

science rules! :)

Original post by Vinchenko

Yeh, the ones you listed are those I am learning as well! :tongue:

Reply 692

appleschnapps

Sorry if I'm being completely blind and missing it, but does anyone have the mark scheme for the specimen paper?

Er, if it exists. That's kind of an assumption on my part.

(edited 12 years ago)

Reply 693

potentialmedicalstudent

Original post by appleschnapps

Sorry if I'm being completely blind and missing it, but does anyone have the mark scheme for the specimen paper?

Er, if it exists. That's kind of an assumption on my part.

it should be on the same document, just scroll down

Reply 694

Schoolio93

Hi can someone exactly tell me what we need to know about neutralisation? Do we need to know how to calculate the enthalpy of dissociation?

Reply 695

appleschnapps

Original post by potentialmedicalstudent

it should be on the same document, just scroll down

Ah, thanks. I had a copy printed out by my teacher, so I didn't think to look at the online version. :smile:

Reply 696

threerose92

Hi can someone please help me to find the molecular formula of a compound that has percentage composition mass: P, 43.7% ; O, 56.3%. The molecular formula mass is 284.

Thanks

Reply 697

wilsea05

Original post by threerose92

Hi can someone please help me to find the molecular formula of a compound that has percentage composition mass: P, 43.7% ; O, 56.3%. The molecular formula mass is 284.

Thanks

43.7/31= 1.41
56.3/16 = 3.52

3.52/1.41 = 2.5
so you have 2.5 Os for every 1 P
which is P2O5 = 31x2+16x5 = 142
284/142 = 2 so times everything by two
so molecular formula will be P4O10

Reply 698

threerose92

Original post by wilsea05

43.7/31= 1.41
56.3/16 = 3.52

3.52/1.41 = 2.5
so you have 2.5 Os for every 1 P
which is P2O5 = 31x2+16x5 = 142
284/142 = 2 so times everything by two
so molecular formula will be P4O10